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3 years ago

Why is the percent increase from 45 to 75 not equal to the percent decrease from 75 to 45? Select three

Mathematics

1 answer:

Novay_Z [31]3 years ago

5 0

Answer:

For me the answer is

The ratio for the percent increase has a smaller denominator than the percent decrease.

Step-by-step explanation:

From 45 to 75, the increase is 30

If we check the % increased, (30/45)*100 = 66.667%

If we decrease the price from 75 to 45, the amount is same, i.e. 30 but the denominator is changed.

(30/75)*100 = 40%

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Mademuasel [1]

306 cause 2448divided by 8

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3 years ago

A set of n = 15 pairs of X and Y scores has SSX = 10, SSY = 40, and SP = 30. What is the slope for the regression equation for p

Mumz [18]

Answer:

Step-by-step explanation:

We have given $SS_X=10$ , $SS_Y=40$ and SP =30

We have to find the slope for regression on Y from X

The slope on Y from X is given by $b_{yx}=\frac{SP}{SS_X}$

As we have given the value of SP =30 and value of $SS_X=10$

So slope for regression will be $b_{yx}=\frac{30}{10}=3$

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3 years ago

I need help!! 6r^(2)-3r-27=-9

Harrizon [31]

Answer:

Step-by-step explanation:

Solution

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3 years ago

At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t

grandymaker [24]

Answer:

(a1) The probability that temperature increase will be less than 20°C is 0.667.

(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.

(b) The probability that at any point of time the temperature increase is potentially dangerous is 0.467.

Step-by-step explanation:

Let X = temperature increase.

The random variable X follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of X is:

$f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.$

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

$P(X$

Thus, the probability that temperature increase will be less than 20°C is 0.667.

(a2)

Compute the probability that temperature increase will be between 20°C and 22°C as follows:

$P(20$

Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.

(b)

Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:

$P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467$

Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c)

Compute the expected value of the uniform random variable X as follows:

$E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5$

Thus, the expected value of the temperature increase is 17.5°C.

7 0

3 years ago

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that 5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

 $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

7 0

3 years ago